Calculus of Real and Complex Variables

11.3 Definitions And Elementary Properties

First is a simple lemma which links an integral to something which is an integer. It will always be the case
that ϕε will denote a mollifier as in Definition 11.2.3. Also, there are various conditions which hold if
something is small enough. Numbers like 1/5 and 1/10 are used to express suitable smallness assumptions.
There is nothing particularly special about these numbers. I just chose some which were sufficiently small
to work out. I am not trying to give anything precise in expressing these estimates. The thing that is
interesting is the degree for continuous functions and anything which will get you this is what is
wanted.

Lemma 11.3.1Lety

∕∈

g

(∂ Ω)

for g ∈ C∞

(-- p)
Ω,ℝ

. Also suppose y is a regular value of g. Then for all εsmall enough,

∫ ∑ { −1 }
ϕε(g(x)− y)detDg (x)dx = sgn(detDg (x)) : x ∈ g (y)
Ω

Proof: First note that the sum is finite. Indeed, if there were a sequence of xk such that xk∈ g−1

(y)

,
then it would have a subsequence converging to some x ∈Ω . Thus g

(x)

= y and by assumption, y

∕∈

∂Ω.
Since y is a regular value, Dg

(x)

is invertible and so by the inverse function theorem, there is an open set
containing x on which g is one to one, contrary to x being a limit of a sequence of points in g−1

(y)

. It only
remains to verify the equation.

I need to show the left side of this equation is constant for ε small enough and equals the right side. By
what was just shown, there are finitely many points,

{xi}

i=1m = g−1

(y )

. By the inverse function theorem,
there exist disjoint open sets Ui with xi∈ Ui, such that g is one to one on Ui with det

(Dg (x))

having
constant sign on Ui and g

(Ui)

is an open set containing y. Then let ε be small enough that
B

(y,ε)

⊆∩i=1mg

(Ui)

and let Vi≡ g−1

(B (y,ε))

∩ Ui.

PICT

Therefore, for any ε this small,

∫ m ∫
ϕ (g (x)− y)detDg (x)dx = ∑ ϕ (g(x)− y)detDg (x)dx
Ω ε i=1 Vi ε

The reason for this is as follows. The integrand on the left is nonzero only if g

(x )

− y ∈ B

(0,ε)

which
occurs only if g

(x)

∈ B

(y,ε)

which is the same as x ∈ g−1

(B(y,ε))

. Therefore, the integrand is nonzero
only if x is contained in exactly one of the disjoint sets, Vi. Now using the change of variables theorem,
(z = g